Question

Two towns $ A $ and $ B $ are connected by a regular bus service with a bus leaving in either direction every $ T $ minutes. A man cycling with a speed of $ 20 \,km\,h^{-1} $ in the direction $ A $ to $ B $ notices that a bus goes past him every $ 18 $ min in the direction of his motion, and every $ 6 $ min in the opposite direction. The period $ T $ of the bus service is

• A. $ 4.5 $ min
• B. $ 9 $ min
• C. $ 12 $ min
• D. $ 24 $ min

Answer 1

The Correct Option is B

Let $ v \,km \,h^{-1} $ be the constant speed with which the bus travel ply between the towns $ A $ and $ B $ . Relative velocity of the bus from $ A $ to $ B $ with respect to the cyclist $ = (v - 20) \,km \,h^{-1} $ Relative velocity of the bus from $ B $ to $ A $ with respect to the cyclist $ = (v + 20) \,km\, h^{-1} $ Distance travelled by the bus in time $ T $ (minutes) $ = vT $ As per question $ \frac{vT}{v-20}=18 $ or $ vT=18v-18\times 2 0 $ and $ \frac{vT}{v+20}=6\quad\ldots\left(i\right) $ or $ vT=6v+20\times6\quad\ldots\left(ii\right) $ Equating $ \left(i\right) $ and $ \left(ii\right) $ , we get $ =18v-18\times20=6v+20\times 6 $ or $ 12v=20\times6+18\times20=480 $ or $ v=40\,km\,h^{-1} $ Putting this value of $ v $ in $ \left(i\right) $ , we get $ 40T=18\times40-18\times20=18\times 20 $ or $ T=\frac{18\times20}{40}=9 $ min